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Question

You swing a bucket of water in a vertical circle at arm's length (.70m). What is the minimum number of revolutions per second you must maintain to keep the water from spilling out of the bucket.

Answer 1

The centrifugal force keeping the water in the bucket is given by (m*v^2)/R, where V is the tangential velocity; this must exceed the weight of the water which is m*g. Then

(m*v^2)/R > m*g

(v^2)/R > g

v^2 > g*R

v > √[g*R]

The tangential velocity is the angular velocity times the radius,

v = w*R (w in radians per second)

one revolution is 2π radians, so

v = 2π*(RPS)*R (RPS = revolutions per second)

RPS = v/(2π*R)

Putting in v = √[g*R] from above

RPS = √[g*R] / 2π*R

RPS = √[g/R] / 2π